Question: You have found the following ages (in years) of all 6 bears at your local zoo: $ 46,\enspace 1,\enspace 20,\enspace 16,\enspace 12,\enspace 2$ What is the average age of the bears at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Explanation: Because we have data for all 6 bears at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $6$ ages and divide by $6$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{6}} x_i}{{6}} $ $ {\mu} = \dfrac{46 + 1 + 20 + 16 + 12 + 2}{{6}} = {16.2\text{ years old}} $ Find the squared deviations from the mean for each bear. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $46$ years $29.8$ years $888.04$ years $^2$ $1$ year $-15.2$ years $231.04$ years $^2$ $20$ years $3.8$ years $14.44$ years $^2$ $16$ years $-0.2$ years $0.04$ years $^2$ $12$ years $-4.2$ years $17.64$ years $^2$ $2$ years $-14.2$ years $201.64$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{888.04} + {231.04} + {14.44} + {0.04} + {17.64} + {201.64}} {{6}} $ $ {\sigma^2} = \dfrac{{1352.84}}{{6}} = {225.47\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{225.47\text{ years}^2}} = {15\text{ years}} $ The average bear at the zoo is 16.2 years old. There is a standard deviation of 15 years.